Find the length of DQ and the angle QDB.
To find the length of DQ, I use cosine rule:
\begin{align}
DQ^2&=QB^2+BD^2-2\cdot QB\cdot BD \cos(\angle QBD)\\
&=9+32-24\sqrt{2}\cos(\angle QBD)\\
&=41-24\sqrt{2}\cos(\angle QBD).
\end{align}
I don't know the angle $\angle QBD$. Now I try to find it using sine rule.
Consider that $\angle TBD =\angle QBD$.
\begin{align}
\dfrac{TD}{\sin(\angle TBD)}=\dfrac{DB}{\sin(\angle DTB)}.
\end{align}
We can see that to find $\angle QBD=\angle TBD$, need the $\angle DTB$. But we don't know the value $\angle DTB$. I'm get stuck here. I can't compute the length DQ and the $\angle QDB$.
Anyone know how to find it?
Best Answer
Use the cosine rule with $\triangle TDB$ to get
$$\begin{equation}\begin{aligned} TD^2 & = TB^2 + BD^2 - 2(TB)(DB)\cos(\measuredangle QBD) \\ 25 & = 25 + 32 - 2(5)(4\sqrt{2})\cos(\measuredangle QBD) \\ -32 & = -40\sqrt{2}\cos(\measuredangle QBD) \\ \cos(\measuredangle QBD) & = \frac{2\sqrt{2}}{5} \end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Alternatively, as suggested by alex's question comment, since $\triangle TDB$ is isosceles (with $TD = TB = 5$), the perpendicular from $T$ bisects $DB$, say at point $E$, so $\triangle TEB$ is right-angled at $E$. Since $EB$ is half of $DB$, i.e., it's $2\sqrt{2}$, this can be used instead to more directly get that
$$\cos(\measuredangle QBD) = \frac{EB}{TB} = \frac{2\sqrt{2}}{5} \tag{2}\label{eq2A}$$
In either case, substituting this into your equation for $DQ^2$ gives
$$\begin{equation}\begin{aligned} DQ^2 & = 41 - 24\sqrt{2}\left(\frac{2\sqrt{2}}{5}\right) \\ & = 41 - \frac{48(2)}{5} \\ & = \frac{205 - 96}{5} \\ & = \frac{109}{5} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Finally, with $DQ = \sqrt{\frac{109}{5}}$, the cosine rule can be used again in $\triangle QDB$ to determine $\cos(\measuredangle QDB)$ and, from that, $\measuredangle QDB$. I'll leave this for you to do.